Q.
The distance of the point (1,0,2) from the point of intersection of the line x−23=y+14=z−212 and the plane x−y+z=16 is
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a
214
b
8
c
321
d
13
answer is D.
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Detailed Solution
x−23=y+14=z−212=tP(3t+2,4t−1,12t+2) lies on x−y+z=16⇒t=1P(5,3,14)A(1,0,2)
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